If $z^2+z+2=0$, then $z^2 + \frac{4}{z^2} = -3$

Question

If $z^2+z+2=0$, for $z\in\mathbb C$, demonstrate: $$z^2 + \frac{4}{z^2} = -3$$

$z^2 = - 2 - z$, but it didn't help me.

Is there any other elegant solution?

Answer 1

Since $z \neq 0$, you have

$$\begin{equation}\begin{aligned} & z^2 + z + 2 = 0 \\ & z + 1 + \frac{2}{z} = 0 \\ & z + \frac{2}{z} = -1 \\ & \left(z + \frac{2}{z}\right)^2 = (-1)^2 \\ & z^2 + 4 + \frac{4}{z^2} = 1 \\ & z^2 + \frac{4}{z^2} = -3 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Answer 2

$z^2+z+2=0$ means $z^2=-z-2$ and $1+z+\dfrac 2z=0$ so $\dfrac2z=-1-z$ so $\dfrac4{z^2}=z^2+2z+1$.

Therefore, $z^2+\dfrac4{z^2}=z^2+z-1=z^2+z+2-3=-3$.

Answer 3

$$z^2+\frac{4}{z^2}+3$$

$$=\frac{z^4+3z^2+4}{z^2}$$

$$=\frac{(z^2+z+2)(z^2-z+2)}{z^2}$$

$$=0$$